We note first that division by zero is not allowed and that for this reason \(2x-4\) may not be equal to zero and that therefore \(x=2\) is not a solution.

We now distinguish two cases, namely \(2x-4>0\) and \(2x-4<0\).
In both cases we multiply the inequality on both sides by \(2x-4\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(2x-4>0\), i.e. \(x> 2\). Then we get \(1<-7(2x-4)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(14x<27\).
Then, dvision by the coefficient of \(x\)gives \(x < {{27}\over{14}}\).
So we have the following system of inequalities: \(x> 2\,\wedge\; x < {{27}\over{14}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(2x-4<0\), i.e. \(x< 2\). Then we get \(1>-7(2x-4)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(14x>27\).
Then, division by the coefficient of \(x\) gives \(x > {{27}\over{14}}\).
So we have the following system of inequalities: \(x< 2\,\wedge\; x > {{27}\over{14}}\)
and this simplifies to \({{27}\over{14}}\lt x\lt 2\).

The solution of the original inequality is \({{27}\over{14}}\lt x\lt 2\).